a stone is thrown vertically upwards with initl vel of 10m/s from a 40m buliding how much time will it.....?

Question

a stone is thrown vertically upwards with initl vel of 10m/s from a 40m buliding how much time will it take to come back to ground

Answer 1

Do this problem in two parts.
First, find how much time it will take to reach its highest point and how high that is.
Then figure the time it will take to fall from that height to the ground.
If the initial velocity was 10m/sec, since the acceleration due to gravity is -9.8m/sec^2 it will take

10/9.8 sec to reach its highest point, or 1.02sec.
Its average velocity for that part of the flight was 10-0/2=5m/s
so it traveled 5.1m.

To find the time it takes to fall from this height of 45.1 m to the ground, use the formula x=1/2 a t^2 where x is the distance the object falls, a is the acceleration due to gravity, and t is the time it takes.

So
45.1=1/2 (9.8m/sec^2)t^2
90.2/9.8=t^2=9.205
t=3.04sec
But the total time for the trip is the time it took to go from the top of the building to its apex, plus the time it took to fall to the ground.
so t(total)=1.02+3.04 sec=4.06 sec.

Answer 2

V - Final velocity.
U - Initial velocity.
a - Acceleration.
t - time taken.
S - Distance.

Time for stone thrown upwards:
V = U + at
0 = 10 + (-9∙81)t
-10 = -9∙81t
t = -10 / -9∙81
t = 1∙019 367 992 sec.

Distance the stone traveled upwards:
V² = U² + 2aS
0² = 10² + 2(-9∙81)S
0² = 100 - 19∙62S
S = -100 / -19∙62
S = 5∙096 839 959 m.

Total distance for stone to travel to the ground from it's maximum height:
Distance = Building height + new height.
Distance = 40 + 5∙096 839 959
Distance = 45∙096 839 959 m

Time to travel from maximum height to ground:
S = Ut + ½at²
45∙096 839 959 = (0)t + ½(-9∙81)t²
45∙096 839 959 = (0)t - 4∙905t²
t² = 45∙096 839 959 / -4∙905
t² = - 9∙194 055 036
t = √( |- 9∙194 055 036)
t = ± 3∙032 170 021

Total time for stone to hit the ground:
Total time = upward time and downward time.
Total time = 1∙019 367 992 + 3∙032 170 021
Total time = 4∙051 538 013 sec.

Answer 3

The formula to this type of a problem is : h = -ut + 1/2 g t^2
From the given data,
40 = -10t + 1/2(10) t^2
40 = -10t + 5 t^2
5t^2 -10t - 40 = 0
5t^2 - 20t + 10t - 40 = 0
5t(t-4) + 10(t-4) = 0
(t-4)(5t+10) = 0
i.e t = 4 or t = -2
Since, time cannot be negative, t = 4
Therefore, time taken by the stone to come back to the ground is 4 seconds.

Answer 4

initially at the same time as the stone is thrown upwards on the launch element the speed will be optimal. by using earth's gravitational rigidity it is about 9.80 one m/sec it is performing in the option course will have a tendency to reduce the speed at which the stone travels. so as a outcome at a level the stone will completely lose its speed and pulled downwards.

Answer 5

If we assume upward to be +ve & downward to be -ve, then

1. Time to reach the building:
s=ut+0.5gt*t, s=0, u=+10, g=(-9.8) gives the desired value of time.
(ignore the value t=0)

2. Time to reach the ground:
s=ut+0.5gt*t, s=+40, u=+10, g=(-9.8) gives desired value of time.
(ignore the negative value).

Answer 6

since u=10 m/s,s=40m, g=-9.8m/s2(negative accelaration) put it in eqn
s= ut + 1/2(a tsquare).
u will get quadratic eqn
sovle for 2 values of time t.

Answer 7

solve: -16t^2 + 10t + 40 = 0

Answer 8

v^2-u^2=2gs
v=o,u=10m/sec,g=9.81,s=?
s=-100/2*9.81=-5.096 m
ignore -ve sign. it indicates against gravity.
now total dist. to come down=40+5.096
=45.096
s=ut+1/gt^2
u=o,s=45.096,g=9.81
t^2=45.096/9.81/2
=45.096*2/9.81=9.19
t=3.03sec